Question

A paper-filled capacitor is charged to a potential difference of V0=2.5 V. The dielectric constant of paper is k=3.7 . The capacitor is then disconnected from the charging circuit and the paper filling is withdrawn, allowing air to fill the gap between the plates. Find the new potential difference V1 of the capacitor. V1= _____ V
While the capacitor is disconnected from the charging circuit, an unknown substance is inserted between the plates. The plates then attain a potential difference that is 0.55 times the original potential difference V0 (when paper filled the capacitor). What is this unknown substance's dielectric constant?

Answer 1

Answer:

k =  6.72

Explanation:

K of paper = 3.7

k of air = 1

Given that charge Q on the capacitor is constant because cell is disconnected from the circuit. So

V = Q / C = 2.5

Capacity becomes C / 3.7 in air .

capacity becomes C/3.7 when paper is replaced by air .

V₁ = Q / (C/3.7)

= 3.7 Q/C

3.7 x 2.5

= 9.25 V

In the second case ,

capacitance  due to new unknown dielectric k

= C/3.7 x k

= kC / 3.7 ( Capacitance in air is C/3.7 )

V ( new ) = Q / ( kC/3.7 )

= 3.7 Q/kC

.55 x 2.5 = 3.7 x( 2.5 / k )

k = 3.7 / .55

= 6.72

Answer 2

The unknown substance's dielectric constant k =  6.72

Calculation of the unknown substance:

We know that

V = Q / C = 2.5

Now

V₁ = Q / (C/3.7)

= 3.7 Q/C

Since V0 = 2.5V

So,

= 3.7 x 2.5

= 9.25 V

Now

In the second case ,

capacitance  because of new unknown dielectric k

So,

= C/3.7 x k

= kC / 3.7 ( Capacitance in air is C/3.7 )

Now

V ( new ) = Q / ( kC/3.7 )

= 3.7 Q/kC

And,

.55 x 2.5 = 3.7 x( 2.5 / k )

k = 3.7 / .55

= 6.72

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